Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is

According to conservation of momentum,$4{\mathrm{mu}}_1=4{\mathrm{mv}}_1+2{\mathrm{mv}}_2$

$$\begin{gathered} \Rightarrow 2\left({\mathrm{u}}_1-{\mathrm{v}}_1\right)={\mathrm{v}}_2 .........\left(\mathrm{i}\right) \\ \mathrm{From} \mathrm{conversation} \mathrm{of} \mathrm{energy}, \\ \frac{1}{2}\left(4\mathrm{m}\right){\mathrm{u}}_1^2=\frac{1}{2}\left(4\mathrm{m}\right){\mathrm{v}}_1^2+\frac{1}{2}\left(2\mathrm{m}\right){\mathrm{v}}_2^2 \\ \Rightarrow 2\left({\mathrm{u}}_1^2-{\mathrm{v}}_1^2\right)={\mathrm{v}}_2^2 ........\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ 2\left({\mathrm{u}}_1^2-{\mathrm{v}}_1^2\right)=4{\left({\mathrm{u}}_1-{\mathrm{v}}_1\right)}^2 ..........\left(\mathrm{iii}\right) \\ \text{ Now, fraction of loss in kinetic energy for mass }4\mathrm{m}\text{ , } \\ \frac{\mathrm{\Delta K}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{{\mathrm{K}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{1/2\left(4\mathrm{m}\right){\mathrm{u}}_1^2-\frac{1}{2}\left(4\mathrm{m}\right){\mathrm{v}}_1^2}{\frac{1}{2}\left(4\mathrm{m}\right){\mathrm{u}}_1^2} .........\left(\mathrm{iv}\right) \\ \text{ Substituting (iii) in (iv), we get }\frac{\mathrm{\Delta K}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{8}{9} \end{gathered}$$